Equation whose solution is a finite tower of $2's$

Question

What equation has a finite power tower of $2's,$ as it's solution:

$$ x=2^{{2^{2}}^{\cdot\cdot\cdot}}.$$

I tried to reverse engineer the solution, back into an equation, so I started with a toy model of the solution:

$x=2^{2^{2}}.$

I took multiple logarithms on both sides and then decided to rest.

Answer 1

I have a simple one, but this is not really exciting.

We have,

$$x = {{{2^{2}}^{2}}^{2}}^{....}$$

Now, I will take logarithms on both sides, just like you did, with base $2$. So,

$$\log_2x = \log_2({{{2^{2}}^{2}}^{2}}^{....})$$

$$\implies \log_2x = {{{2^{2}}^{2}}^{2}}^{....} \text{[The power tower is infinite.]}$$

$$\implies\log_2 x = x \implies 2^x = x$$

That's it.

Wolfram Alpha also shows an appropriate solution expressed with Lambert W Function:

$$x = \frac{W_n(-\log(2))}{\log(2)}, \text{where n} \in \Bbb{Z}$$.

Eisenstein confirmed a Power Tower can be expressed in the form :

$$ {{{a^{a}}^{a}}^{a}}^{....} = \frac{W_n(-\log(a))}{\log(a)}, \text{where n} \in \Bbb{Z}$$

So The Solution of WolframAlpha confirms $$x = {{{2^{2}}^{2}}^{2}}^{....}$$

Source : Lambert W Function : WolframAlpha

Answer 2

$$ x - 2^{{2^{2}}^{\cdot\cdot\cdot}} = 0.$$

Answer 3

I feel that the easiest equation that you can find for this is quite beautiful actually.

Consider, $$ x = 2^x $$ This is based on the simple observation, which is quite useful in such infinite series, that there is an exact copy of x in the exponential of 2. This recursive observation would lead you to writing the RHS as $ 2^x$.

Answer 4

Solving this equation for $x$ gives the desired solution:

$f(x)=2^{(2^2A)}=\frac{1}{A};$ $A=\frac{1}{\log_2(\log_2(\log_2(x)))}.$

Here the nested logarithms can be nested further to give an arbitrarily high power tower.